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$X_{2} (g) + Y_{2} (g) ⇌ 2 X Y (g)$ reaction was studied at a certain temperature. In the beginning, 1 mole of $x_{2}$ was taken in a one litre flask and 2 moles of $Y_{2}$ was taken in another 2 litre flask and both these containers are connected so that equilibrium can be established. What is the equilibrium concentration of $X_{2}$ and $Y_{2}$ ? Given Equilibrium concentration of $[X Y] = 0.6 moles / litre$ .

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$(\frac{1}{3} - 0.6), (\frac{2}{3} - 0.6)$

$(1 - 0.3), (2 - 0.3)$

$(\frac{1}{3} - 0.3), (\frac{2}{3} - 0.3)$

$(1 - 0.6), (2 - 0.6)$

answer is A.

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Detailed Solution

$X_{2} + Y_{2} ⇌ 2 X Y$

$\frac{1}{3} - x$ $\frac{2}{3} - x$ $2 x = 0.6$

$\Rightarrow x = 0.3$

$= (\frac{1}{3} - 0 .3), (\frac{2}{3} - 0 .3)$

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